Question

Total number of possible isomers of $\left[Cu\right({NH}_3{)}_4]\left[{PtCl}_4\right]$ is:

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

4

The total number of possible isomer of the $\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4]$ is 4.

The isomers can be formed by exchanging the ligands.

1.$\left[Cu\right(N{H}_3{)}_{3}Cl\left]\right[Pt\left(Cl{)}_{3}N{H}_3\right]$

2.$\left[Cu\right(N{H}_3{)}_{2}C{l}_2\left]\right[PtC{l}_2\left(N{H}_3{)}_2\right]$

3.$\left[Cu\right(N{H}_3\left)C{l}_3\right]\left[PtCl\right(N{H}_3{)}_3]$

4. $\left[CuC{l}_4\right]\left[Pt\right(N{H}_3{)}_4]$

Hence, the correct option is B